Troubleshooting Rolling Log Acceleration on a Ramp

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SUMMARY

The discussion focuses on calculating the linear acceleration of a uniform solid cylindrical log rolling down a ramp inclined at 28 degrees. The initial calculation incorrectly used the vertical height as 4.20 m instead of the actual height derived from the ramp's angle. The correct height should be calculated as 4.20 m multiplied by the sine of 28 degrees, which resolves the discrepancy in the acceleration calculation. The final correct linear acceleration can be derived using the adjusted height.

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Mdhiggenz
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Homework Statement



A uniform solid cylindrical log begins rolling without slipping down a ramp that rises 28.0 above the horizontal. After it has rolled 4.20 m along the ramp, the magnitude of its linear acceleration is closest to

Homework Equations





The Attempt at a Solution



What I did was

mgh=1/2mv^2+1/2Iω^2

mgh=1/2mv^2+1/2[1/2mr^2]ω^2

v=rω

mgh=1/2mv^2+1/4mv^2
√4/3gh=v

solved for v and got 7.41m/s

I then used the constant acceleration equations

vx^2=vo+2aΔx

solving for a and got 6.54 which is the incorrect answer.

Where did I go wrong

Thank you!
 
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Hi Mdhiggenz,

The problem is in your first calculation, where you come up with 7.41m/s. Your manipulation of the algebra looks correct to me (double check!), but at the very last step you seem to have used h=4.20m. The problem says that the log moved 4.20m along the ramp, which means it would have descended 4.20m*sin(28deg), which is the value you should use for h. Does this fix your problem?

Hope this helps,
Bill Mills
 
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